Finite Fourier Transform Let $u=\lbrace u_0,\ldots,u_{N-1}\rbrace$. Then the finite discrete Fourier transform of $s$ is given by: $$ \hat u_k = \sum_{n=0}^{N-1} u_n e^{-i\frac{2\pi}{N} kn}\ . $$
Plancherel-Parseval Let $u=\lbrace u_0,\ldots,u_{N-1}\rbrace$ and let $\hat u$ its Fourier transform. Then $$ \Vert u\Vert_2^2 = \frac{1}{N}\Vert \hat u\Vert_2^2\ , $$ ie. $$ \sum_{n=0}^{N-1} |u_n|^2 = \frac{1}{N}\sum_{k=0}^{N-1}|\hat u_k|^2 . $$ Let $v=\lbrace v_0,\ldots,v_{N-1}\rbrace$ and let $\hat v$ its Fourier transform. Then $$ \langle u,v \rangle = \frac{1}{N}\langle \hat u,\hat v \rangle $$ ie. $$ \sum_{n=0}^{N-1} u_nv_n = \frac{1}{N} \sum_{k=0}^{N-1} \hat u_k \hat v_k \ . $$
Inverse Finite Fourier Transform Let $u=\lbrace u_0,\ldots,u_{N-1}\rbrace$. Let $\hat u$ its Fourier transform. Then $$ u_n = \frac{1}{N} \sum_{k=0}^{N-1} \hat u_k e^{i\frac{2\pi}{N} kn}\ . $$
For the calculus properties, the finite signals $u$ and $v$ must be both considered as $N$-periodic discrete signal, by using $0$ padding if necessary. The used convolution is then the circular convolution.
Convolution: $$\widehat{u* v}[k] = \hat u[k] \hat v[k]$$ Multiplication: $$\widehat{u . v}[k] = (\hat u * \hat v ) [k]$$ Translation: let $v_a[n] = u[n-a]$ $$\hat v_a[k] = e^{-i\frac{2\pi}{N} ka} \hat u[k]$$ Modulation: let $v\_{\theta}[n] = e^{i2\pi \theta t}u[n]$ $$\hat{v}_\theta(\nu) = \hat u(\nu - \theta)$$ Complexe conjuguée $$\widehat{\bar u}[k] = \overline{ \hat{ u}[N-k]}$$ Symétrie hermitienne $$u[n]\in\mathbb{R} \Rightarrow \hat u[N-k] = \overline{\hat{ u}[k]}$$